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Find the number of solutions to

x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_10 <= 2

in nonnegagive integers.

 Dec 30, 2022
 #1
avatar+23227 
0

Counting the ways:

 

1) You can have 10 zeros ==  1 way

 

2) You can have 1 one and 9 zeros == 10 ways

 

3) You can have 1 two and 9 zeros == 10 ways

 

4) You can have 2 ones and 8 zeros:

     If you start with a 1: 9 ways to place the other one.

     If you start with 0 1: 8 ways to place the other one.

     If you start with 0 0 1: 7 ways to place the other one.

     If you start with 0 0 0 1: 6 ways to place the other one.

      

     ... continue with this and add all the ways.

 Dec 30, 2022
 #3
avatar+28 
0

I like your solution, for a second I thought you got it wrong nvm

hairyberry  Dec 30, 2022
edited by hairyberry  Dec 30, 2022
 #2
avatar+28 
+1

I agree with your solution, instead I did this another way

I started with considering all the possible numbers this could be equal to because of the inequality sign, 0 1 or 2

For zero there can only be 1 possibility, all zeros

For x_1+x_2...x_10 equal to 1, there are 10 possibilities, one of them is one, the others are zero 

For x_1+x_2...x+10 to be equal to 2, we use stars and bars (we could have done for the first two cases, but I'm too lazy cool), so it would be (10+2-1) choose (10-1) which is 11 choose 9 which is 55

Adding all of these together we get 55+10+1 which is 66

 Dec 30, 2022

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